Question

for exercises 2–5, find the values of x and y.
2.
3.
4.
5.

Explanation:

Exercise 2

Step1: Find hypotenuse length

The hypotenuse of the large right triangle is $6 + 14 = 20$.

Step2: Solve for $x$ (leg)

Use geometric mean theorem: $x^2 = 6 \times 20$
$x = \sqrt{120} = 2\sqrt{30}$

Step3: Solve for $y$ (altitude)

Use geometric mean theorem: $y^2 = 6 \times 14$
$y = \sqrt{84} = 2\sqrt{21}$

Exercise 3

Step1: Find hypotenuse length

The hypotenuse of the large right triangle is $1 + 9 = 10$.

Step2: Solve for $x$ (leg)

Use geometric mean theorem: $x^2 = 1 \times 10$
$x = \sqrt{10}$

Step3: Solve for $y$ (leg)

Use geometric mean theorem: $y^2 = 9 \times 10$
$y = \sqrt{90} = 3\sqrt{10}$

Exercise 4

Step1: Split hypotenuse into segments

Let the left segment be $12 - x$. Use geometric mean theorem: $4^2 = (12 - x) \times 12$
$16 = 144 - 12x$

Step2: Solve for $x$

$12x = 144 - 16 = 128$
$x = \frac{128}{12} = \frac{32}{3}$

Step3: Solve for altitude (let it be $h$)

$h^2 = (12 - \frac{32}{3}) \times \frac{32}{3} = \frac{4}{3} \times \frac{32}{3}$
$h = \frac{8\sqrt{2}}{3}$

Exercise 5

Step1: Find hypotenuse length

The hypotenuse of the large right triangle is $36$.

Step2: Solve for $x$ (other leg)

Use Pythagorean theorem: $x^2 + 12^2 = 36^2$
$x = \sqrt{1296 - 144} = \sqrt{1152} = 24\sqrt{2}$

Step3: Solve for $y$ (altitude)

Use area equality: $\frac{1}{2} \times 12 \times 24\sqrt{2} = \frac{1}{2} \times 36 \times y$
$y = \frac{12 \times 24\sqrt{2}}{36} = 8\sqrt{2}$

Answer:

1. $x=2\sqrt{30}$, $y=2\sqrt{21}$
2. $x=\sqrt{10}$, $y=3\sqrt{10}$
3. $x=\frac{32}{3}$, altitude $=\frac{8\sqrt{2}}{3}$
4. $x=24\sqrt{2}$, $y=8\sqrt{2}$